Characterization of Dual Operator

Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ and $Y$ be normed vector spaces over $\GF$.

Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.

Let $X^{\ast \ast}$ and $Y^{\ast \ast}$ be the second normed duals of $X$ and $Y$ respectively.

Let $T : Y^\ast \to X^\ast$ be a bounded linear transformation.

Let $\iota_X : X \to X^{\ast \ast}$ be the evaluation linear transformation for $X$.

Let $\iota_Y : Y \to Y^{\ast \ast}$ be the evaluation linear transformation for $Y$.

The following statements are equivalent:

$(1) \quad$ $T$ is $\struct {w^\ast, w^\ast}$-continuous

$(2) \quad$ there exists a bounded linear transformation $S : X \to Y$ such that $S^\ast = T$, where $S^\ast$ denotes the dual operator of $S$

$(3) \quad$ $T^\ast \sqbrk {\iota_X X} \subseteq \iota_Y Y$

Proof

$(1)$ implies $(2)$

Suppose that $T$ is $\struct {w^\ast, w^\ast}$-continuous.

From Characterization of Continuity of Linear Functional in Weak-* Topology ‎, we have that:

$x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.

From Composite of Continuous Mappings is Continuous:

$x^\wedge \circ T : \struct {Y^\ast, w^\ast} \to \GF$ is continuous.

Applying Characterization of Continuity of Linear Functional in Weak-* Topology , there exists $S x \in Y$ such that:

$x^\wedge \circ T = \paren {S x}^\wedge$

We first show that $S : X \to Y$ is linear.

We have:

\(\ds \paren {\map S {\alpha x + \beta y} }^\wedge\)

\(=\)

\(\ds \paren {\alpha x + \beta y}^\wedge \circ T\)

\(\ds \)

\(=\)

\(\ds \alpha \paren {x^\wedge \circ T} + \beta \paren {y^\wedge \circ T}\)

Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual

\(\ds \)

\(=\)

\(\ds \alpha S x + \beta S y\)

We now show that $S$ is bounded.

We have, for each $x \in X$:

\(\ds \norm {S x}_Y\)

\(=\)

\(\ds \norm {\paren {S x}^\wedge}_{Y^{\ast \ast} }\)

\(\ds \)

\(=\)

\(\ds \norm {x^\wedge \circ T}_{Y^{\ast \ast} }\)

\(\ds \)

\(\le\)

\(\ds \norm {x^\wedge}_{X^{\ast \ast} } \norm T_{\map B {Y^\ast, X^\ast} }\)

Norm on Bounded Linear Transformation is Submultiplicative

\(\ds \)

\(=\)

\(\ds \norm T_{\map B {Y^\ast, X^\ast} } \norm x_X\)

Evaluation Linear Transformation on Normed Vector Space is Linear Isometry

So $S$ is bounded.

Finally, for $f \in Y^\ast$ and $x \in X$ we have:

\(\ds \map f {S x}\)

\(=\)

\(\ds \map {\paren {S x}^\wedge} f\)

\(\ds \)

\(=\)

\(\ds \map {\paren {x^\wedge \circ T} } f\)

\(\ds \)

\(=\)

\(\ds \map {x^\wedge} {T f}\)

\(\ds \)

\(=\)

\(\ds \map {\paren {T f} } x\)

Definition of Evaluation Linear Transformation on Normed Vector Space

That is:

$f \circ S = T f$

for each $f \in Y^\ast$.

From the definition of the dual operator we have:

$S^\ast f = T f$

for each $f \in Y^\ast$.

So $S^\ast = T$.

$\Box$

$(2)$ implies $(3)$

Suppose that:

there exists a bounded linear transformation $S : X \to Y$ such that $S^\ast = T$, where $S^\ast$ denotes the dual operator of $S$.

We have $S^{\ast \ast} = T^\ast$, where $S^{\ast \ast}$ denotes the second dual operator.

Let $x \in X$ and $f \in Y^\ast$.

Then:

\(\ds \map {\map {T^\ast} {x^\wedge} } f\)

\(=\)

\(\ds \map {\map {S^{\ast \ast} } {x^\wedge} } f\)

\(\ds \)

\(=\)

\(\ds \map {\paren {x^\wedge \circ S^\ast} } f\)

Definition of Dual Operator

\(\ds \)

\(=\)

\(\ds \map {x^\wedge} {S^\ast f}\)

\(\ds \)

\(=\)

\(\ds \map {x^\wedge} {f \circ S}\)

Definition of Dual Operator

\(\ds \)

\(=\)

\(\ds \map f {S x}\)

Definition of Evaluation Linear Transformation on Normed Vector Space

\(\ds \)

\(=\)

\(\ds \map {\paren {S x}^\wedge} f\)

Definition of Evaluation Linear Transformation on Normed Vector Space

That is:

$\map {T^\ast} {\iota_X x} = \map {\iota_Y} {S x}$

$\Box$

$(3)$ implies $(1)$

Suppose that $T^\ast \sqbrk {\iota_X X} \subseteq \iota_Y Y$.

For each $x^\wedge \in \iota_X X$, we then have that:

$\map {T^\ast} {x^\wedge} \in \iota_Y Y$

That is, there exists $y \in Y$ such that:

$x^\wedge \circ T = y^\wedge$

From Characterization of Continuity of Linear Functional in Weak-* Topology, we then have:

$x^\wedge \circ T : \struct {Y^\ast, w^\ast} \to \GF$ is continuous

for each $x \in X$.

Conversely from Characterization of Continuity of Linear Functional in Weak-* Topology, we have that every continuous linear functional $\Phi : \struct {Y^\ast, w^\ast} \to \GF$ has the form $\Phi = x^\wedge$.

So from Continuity in Initial Topology, we have that:

$T : \struct {Y^\ast, w^\ast} \to \struct {X^\ast, w^\ast}$ is continuous.

$\blacksquare$